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Topic: f(z) + abs(z) = 0, where z is complex
Replies: 18   Last Post: Jul 11, 2011 9:23 AM

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slawek

Posts: 25
Registered: 12/12/08
Re: f(z) + abs(z) = 0, where z is complex
Posted: Jul 11, 2011 2:10 AM
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U?ytkownik "Gib Bogle" napisa? w wiadomo?ci grup
dyskusyjnych:ivdv92$384$2@speranza.aioe.org...

>If you can't find an analytical solution to your pair of equations then you
>might be out of luck.


1. Analytical solutions are not avaliable.

2. The good approximate values are given "for free".

3. The f(z) is not a simple one. Indeed, the problem was Sum[ a(z,n,m) +
b(z,n,m) abs(z)^2 = c(n,m) , n = 1, ..., N ] where m = 1, ..., M .

4. FYI: There is a trouble "what is a solution?" The eq. -1 + abs(x) = 0,
where x is a real variable, have got two dinstinct solutions x1 = -1 and x2
= +1. The eq. -1 + abs(z) = 0, where z is complex, have got the circle as
the solution. Therefore it is possible to obtain a continuum instead a
single number. But the next routine need a number, not a functional
dependency.

slawek






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